Four resistors each of 6\ Ω are connected in various arrangements as…

The correct answer is Option (3) → (C), (A), (B), (D)

Use \( R_{\mathrm{s}} = R_{1} + R_{2} + \ldots \) for resistors in series and \( 1/R_{\mathrm{p}} = 1/R_{1} + 1/R_{2} + \ldots \) for resistors in parallel. Read statement (B) in the same way as statement (A): a parallel combination of [three resistors in series] with the fourth resistor.

(A) A parallel combination of two resistors in series with the combination of other two resistors in series. — Two resistors in series give \( 12\ \Omega \); two such branches in parallel:

    \( \displaystyle R_{A} = \frac{12 \times 12}{12 + 12} = 6\ \Omega \)

(B) A parallel combination of three resistors in series with the fourth resistor. — Three resistors in series give \( 18\ \Omega \); this in parallel with the fourth \( 6\ \Omega \) resistor:

    \( \displaystyle R_{B} = \frac{18 \times 6}{18 + 6} = 4.5\ \Omega \)

(C) All four resistors are in series. —

    \( R_{C} = 4 \times 6 = 24\ \Omega \)

(D) All four resistors are in parallel. —

    \( R_{D} = 6/4 = 1.5\ \Omega \)

Decreasing order: \( 24 \gt 6 \gt 4.5 \gt 1.5\ \Omega \), that is, (C), (A), (B), (D).

Note: Statement (B) does not say how the words are to be grouped. If it is read as [three resistors in parallel] in series with the fourth resistor, then \( R_{B} = 6/3 + 6 = 2 + 6 = 8\ \Omega \), and the decreasing order becomes \( 24 \gt 8 \gt 6 \gt 1.5\ \Omega \), that is, (C), (B), (A), (D). Under that reading Option (1) would be correct. Reading statement (B) in the same way as statement (A), as done above, gives the keyed Option (3).